Solve Prob, 10.89, assuming that the load is applied at a point 40 mm from the geometric axis an

step 1 We start by drawing the free body diagram as seen in the figure. Next step is to use equation wh

Solve Prob, 10.89, assuming that the load is applied at a...

Key Concepts

Eccentric Loading

Eccentric loading occurs when a force is applied at a point that is offset from the centroidal or geometric axis of a structural member. This misalignment introduces bending moments in addition to direct axial loads, which can modify the stress distribution within the member and significantly impact its performance under load, often requiring special analysis in design to ensure stability and safety.

Effective Length

Effective length is a concept used in stability and buckling analysis to represent the equivalent length of a column or structural element considering its end constraints. Instead of using the actual physical length, an effective length is calculated based on boundary conditions, and it is used in formulas to predict critical loads and buckling behavior, ensuring that the influence of different support conditions is appropriately accounted for.

Combined Axial and Bending Effects

When a load is applied eccentrically, the structural element experiences both axial compression and bending moments. This combination, often referred to as combined loading, requires analysis that simultaneously considers both effects to accurately assess stresses and deformations. In design and analysis, combining these effects is critical to avoid failures that might arise from the amplified stresses due to the interaction of axial and bending loads.

Transcript

00:01 We start by drawing the free body diagram as seen in the figure.

00:05 Next step is to use equation which state that for a static problem applied to this case, it is true that sigma f is equal to zero.

00:26 And so we can write sigma m o is equal to sigma r cross f is equal to zero.

00:39 Now let's stop here for a moment and think the force equal to the force equal to the force.

00:44 Equation will provide three equations that will have five unknowns.

00:49 Three forces at c, tensions at de and ad.

00:54 The momentum equilibrium, however, taken around c removes it.

00:59 So we can use that to find the tensions de and ad.

01:05 So let's write all distances from c of the three points.

01:10 The distance from c to c is zero, at which forces are applied since we will need them when writing the moment.

01:20 So, accordingly, i can write rbc is equal to 0i plus 0j plus 2 .4k.

01:39 Rdc is equal to minus 0 .8i plus 0 .6j plus 0k.

01:49 And rec is equal to 0 .8i plus 1 .2j plus 0k.

02:03 We have to do one more thing before moving on.

02:07 Before i said the tension at ad and ed has only one unknown, even though it is a three -dimensional force.

02:17 The reason for this is because we know that the tensions ad and a .e.

02:22 Will be in the direction of vectors ad and ae, since that's the way cables behave.

02:31 If we find the unit vector in this directions, we can write f -a -d is equal to f -a -d -u -a -d, and f -a -e is equal to f -a -e -u -e, u -a -e, where the unit vectors uad and uae can be returned as or can be found as uad is equal to ad upon magnitude of ad and uae is equal to a .e is equal to a .e upon magnitude of a .e.

03:32 The vectors ad and ae can be found from the free body diagram.

03:40 So, these are ad is equal to minus 0 .8i plus 0 .6j minus 2 .4k.

03:57 And ae is equal to 0 .8i plus 1 .2j minus 2 .4k.

04:10 And the lengths are or the magnitudes are found as ad is equal to, as we know, it can be calculated as square root of 0 .8 square plus 0 .6 square plus minus 2 .4 square, which gives the answer 2 .6 meter.

04:38 And the magnitude of a will be calculated as square root of 0 .8 square plus 1 .2 square plus minus 2 .4 square gives the answer 2 .8 meter.

04:55 So, this becomes f.

05:02 A .d.

05:04 U .a .d.

05:05 Is equal to minus 0 .8 upon 2 .6i plus 0 .6 upon 2 .6.

05:17 X minus 2 .4 upon 2 .6 k times f a .d and f a .e u a d is equal to 0 .8 upon 2 .8 i plus 1 .2 .8 j minus 2 .8k times f a d now we can move on to the moment equilibrium equation around c which is sigma mc is equal to sigma r cross f is equal to zero so here we are going to take the cross product of r and f which will be r c cross c plus r dc cross f a d c plus r c plus r c cross f a plus r c cross f a plus r c plus r -c -b -cross -b will be 0.

06:41 And in short, by simplifying this terms, i can write r -d -c -cross -f -a -d plus r -e -c -cross -f -a -e plus r -c -b -c -b will be equivalent to 0.

07:11 Now we find each of these products individually.

07:19 So first of all, we will take the cross product of this too.

07:23 Then we will solve for this one and then we will solve for this one.

07:28 So let's solve the first cross product that is rdc cross fad.

07:36 Yes, this is the cross product...

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